Your steps were correct, but Bernoulli's inequality will not help you here.

[tex]\frac{n}{6^n}\leq \frac{n}{1+5n}[/tex]

Is correct, but you can't use it, since the right hand side obviously does not approach zero.
It's basically the squeeze theorem. If you want to do it by comparison, you need something bigger than [itex]\frac{n}{6^n}[/itex] that will still converge to 0.

I think quasar987 was trying to prove the convergence on a sequence not on series.
instead of writing (1+5)^n, it's better (5+1)^n or 6^n>=
>= 5^n+n5^(n-1)+... or conveniently 6^n>n5^(n-1) because the term n5^(n-1) in closer to 6^n than 1+5n.
finally=> (n/6^n)<(n/n5^(n-1)=(1/5^(n-1)< e =>
n>=((lne^-1)/ln5) +1

But if the last line of my first post is correct, it is precisely saying I have found that N in the person of

[tex]\frac{\epsilon}{1-5\epsilon}[/tex]

, no?

Galileo said:

It's basically the squeeze theorem. If you want to do it by comparison, you need something bigger than [itex]\frac{n}{6^n}[/itex] that will still converge to 0.

So are you saying that whenever I use the argument of saying "this is smaller than that, therefor if I find the N for "that", it will be valid for "this" too", I have first to verify that "that" has the same limit as I'm trying to show "this" has? The limit evaluating process would never end!

Mmh, thanks for pointing that out Galileo. I hadn't noticed this is what we were doing even when we prove that a limit is something different then 0.

Muzza, sorry about that, I thought your post #8 was only meant to show that my N fails for epsilon < 1/5. However, I don't understand what do you mean by

Muzza said:

But if [itex]\epsilon < 1/5[/itex], then [itex]1 - 5 \epsilon > 0[/itex], so that

[tex]
n < \frac{\epsilon}{1 - 5 \epsilon}.
[/tex]

What does it matter what the value of [itex]1 - 5 \epsilon > 0[/itex] is? Starting from the hypothesis expression that concerns any epsilon at all, you have arrived at an equivalent form, that is [itex]n(1 - 5 \epsilon) < \epsilon[/itex]. Isn't the natural and valid for all epsilon (except 1/5) next step

[tex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/tex]?!

However, I do realise that the very fact that [itex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/itex] explodes in my face for epsilon = 1/5 implies that the hypothesis expression is wrong for epsilon = 1/5, and therfor, wrong overall. And also, for epsilon < 1/5, [itex]n < \frac{\epsilon}{1 - 5 \epsilon}.[/itex] is true only for n < 1, therefor, for no n, since n is natural and so there exists no N. (perhaps this is what you meant?)

There is one last thing that bothers me though... starting from [itex]n < \epsilon + 5 \epsilon n[/itex], I proceeded in a different way, and I don't see that's wrong with it..

Aren't you familiar with how multiplication by a negative number affects an inequality? As in, if you multiply (or divide (same thing)) by a negative number, you must switch the inequality sign, if you multiply by a positive number you can leave it alone?

If 1 - 5e > 0, the inverse of (1 - 5e) is also positive, so one can multiply both sides of n(1 - 5e) < e by 1/(1 - 5e) to get n < e/(1 - 5e).

But if (1 - 5e) < 0, you must switch the signs around, so you get n > e/(1 - 5e).